Question

For each of the matricesAin Exercises 7 through 11, find an orthogonal matrix S and a diagonal matrix Dsuch that S^-1AS. Do not use technology.

10. $A=\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]$

Short answer

The diagonal matrix is $D=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 9\end{array}\right]$ and the orthogonal matrix is $S=\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right]$.

Step-by-step solution

Define symmetric matrix

• In linear algebra, a symmetric matrix is a square matrix that does not change when its transpose is calculated.
• A symmetric matrix is defined as one whose transpose is identical to the matrix itself.
• A square matrix of size n x nis symmetric if B^T=B.

Find the eigenvalues of the given matrix

Given,

$A=\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]$

According to theorem 7.2.1,

role="math" localid="1668575473197" $\det \left(A-\lambda I_n\right)=0$

The formula for finding eigenspace is

$E_{\lambda }=\ker \left(A-\lambda I_n\right)=\left\{\overrightarrow{v}\text{in}{\mathrm{ℝ}}^n:A\overline{v}=\lambda \overline{v}\right\}$

Now we need to find an orthonormal eigenbasis for the given matrix A.

$$\begin{gathered} \Rightarrow det\left(\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]-\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]\right)=0 \\ \Rightarrow det\left(\left[\begin{array}{ccc}1-\lambda & -2& 2\\ -2& 4-\lambda & -4\\ 2& -4& 4-\lambda \end{array}\right]\right)=0 \\ \Rightarrow (1-\lambda )\left[{(4-\lambda )}^2-16\right]+2[-2(4-\lambda )+8]+2[8-2(4-\lambda \left)\right]=0 \\ \Rightarrow (1-\lambda )\left[16+{\lambda }^2-8\lambda -16\right]+2[-8+2\lambda +8]+2[8-8+2\lambda ]=0 \\ \Rightarrow {\lambda }^2-8\lambda -{\lambda }^3+8{\lambda }^2+4\lambda +4\lambda =0 \\ \Rightarrow -{\lambda }^3+9{\lambda }^2=0 \\ \Rightarrow -{\lambda }^2(\lambda -9)=0 \\ \Rightarrow \lambda =0,0,9 \end{gathered}$$

The eigenvalues of the given matrix A are 0,0,9.

Find the eigenspaces for λ=0

Eigenspace for$\lambda =0$

$$\begin{gathered} E_0=\ker \left(A-0I_3\right) \\ =ker\left(\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\right)=0 \\ =ker\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right] \end{gathered}$$

Now find the kernel of the matrix.

$\Rightarrow \left[\begin{array}{ccc}1& -2& 2\\ -2& 1& -4\\ 2& -4& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new third row:

$\left(R_3-2R_1\to R_3\right)\Rightarrow \left[\begin{array}{ccc}1& -2& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

New second row:

$\left(R_2+2R_1\to R_2\right)\Rightarrow \left[\begin{array}{ccc}1& -2& 2\\ 0& 0& 0\\ 2& -4& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

From the above equation we get the solution as

$$\begin{gathered} x_1-2x_2+2x_3=0 \\ \Rightarrow x_1=2x_2-2x_3 \end{gathered}$$

Therefore, the eigenvector of the given matrix can be represented as:

$$\begin{gathered} \Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}2x_2-2x_3\\ x_2\\ x_3\end{array}\right] \\ =\left[\begin{array}{c}2x_2\\ x_2\\ 0\end{array}\right]+\left[\begin{array}{c}-2x_3\\ 0\\ x_3\end{array}\right] \\ =x_2\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]-x_3\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right],x_2,x_3\in R \end{gathered}$$

Hence the eigenspace for $\lambda =0$is $E_0=span\left(\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]\right)$.

Find the eigenspaces for λ=9

Eigenspace for$\lambda =9$

$$\begin{gathered} E_0=\ker \left(A-9I_3\right) \\ =ker\left(\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]-\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]\right)=0 \\ =ker\left[\begin{array}{ccc}-8& -2& 2\\ -2& -5& -4\\ 2& -4& -5\end{array}\right] \end{gathered}$$

Find the kernel of the matrix.

$=ker\left[\begin{array}{ccc}-8& -2& 2\\ -2& -5& -4\\ 2& -4& -5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new first row.

role="math" localid="1668579108489" $$\begin{gathered} \left(-\frac{1}{8}\times R_1\to R_1\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ -2& -5& -4\\ 2& -4& -5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left(R_1-\frac{1}{4}{R}_2\to R_1\right)=ker\left[\begin{array}{ccc}1& 0& -\frac{1}{2}\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

To get new third row:

$$\begin{gathered} \left(R_3-2R_1\to R_3\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ 0& -\frac{9}{2}& -\frac{9}{2}\\ 0& \frac{9}{2}& -\frac{9}{2}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left(R_3+\frac{9}{2}{R}_2\to R_3\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

New second row:

$$\begin{gathered} \left(R_2+2R_1\to R_2\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ 0& -\frac{9}{2}& -\frac{9}{2}\\ 2& -4& -5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left(\frac{1}{5}\times R_2\to R_2\right)=ker\left[\begin{array}{ccc}1& \frac{1}{4}& -\frac{1}{4}\\ 0& 1& 1\\ 0& -\frac{9}{2}& -\frac{9}{2}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

From equation above we get,

$$\begin{gathered} x_1+0-\frac{1}{2}{x}_3=0\Rightarrow x_1=\frac{1}{2}{x}_3 \\ 0+x_2+x_3=0 \end{gathered}$$

Therefore the eigenvector is given as

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}{x}_3\\ -x_3\\ x_3\end{array}\right]=\frac{1}{2}{x}_3\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right],x_3\in R$

Hence the eigenspace for $\lambda =9$is $E_9=span\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$.

Find the orthogonal matrix

The eigenspaces are perpendicular to each other. The orthonormal eigenbasis can be calculated as follows:

$$\begin{gathered} \overrightarrow{v_1}=\frac{1}{\sqrt{2^2+1^2+0^2}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right] \\ \overrightarrow{v_1}=\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right] \\ \overrightarrow{v_2}=\frac{\left(\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]-\left({\displaystyle \frac{1}{\sqrt{5}}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right].\left[\begin{array}{c}2\\ 0\\ -1\end{array}\right]\right)\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]\right)}{\left|\left(\left[\begin{array}{c}2\\ -1\end{array}\right]-\left[\begin{array}{c}2\\ \sqrt{5}\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\right)\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]\right|} \end{gathered}$$

$$\begin{gathered} =\frac{\sqrt{5}}{3}\left[\begin{array}{c}\frac{2}{5}\\ -\frac{4}{5}\\ -1\end{array}\right] \\ =\overrightarrow{v_2}\left[\begin{array}{c}\frac{2}{3\sqrt{5}}\\ -\frac{4}{3\sqrt{5}}\\ \overrightarrow{v_3}\end{array}\right] \end{gathered}$$

$$\begin{gathered} \overrightarrow{v_3}=\frac{5}{3}\left[-\frac{1}{3\sqrt{5}}\right]\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right] \\ S=\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right] \end{gathered}$$

Find the inverse of orthogonal matrix:

$$\begin{gathered} S^{-1}={\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right]}^{-1} \\ =\frac{1}{\left({\displaystyle \frac{20}{27}}\right)}adj\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right] \\ \Rightarrow S^{-1}=S=\left[\begin{array}{ccc}-\frac{27}{10\sqrt{5}}& -\frac{9}{10\sqrt{5}}& -\frac{9}{20}\\ -\frac{27}{20\sqrt{5}}& \frac{9}{5\sqrt{5}}& \frac{9}{10}\\ 0& \frac{9\sqrt{5}}{20}& -\frac{9}{10}\end{array}\right] \end{gathered}$$

Hence the orthogonal matrix is $S=\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right]$.

Find the diagonal matrix

Now we have to find the diagonal matrix as below:

$$\begin{gathered} D=S^{-1}AS \\ \left[\begin{array}{ccc}-\frac{27}{10\sqrt{5}}& -\frac{9}{10\sqrt{5}}& -\frac{9}{20}\\ -\frac{27}{20\sqrt{5}}& \frac{9}{5\sqrt{5}}& \frac{9}{10}\\ 0& \frac{9\sqrt{5}}{20}& -\frac{9}{10}\end{array}\right]\left[\begin{array}{ccc}1& -2& 2\\ -2& 4& -4\\ 2& -4& 4\end{array}\right]\left[\begin{array}{ccc}\frac{2}{\sqrt{5}}& \frac{2}{3\sqrt{5}}& \frac{1}{3}\\ \frac{1}{\sqrt{5}}& -\frac{4}{3\sqrt{5}}& -\frac{2}{3}\\ 0& -\frac{5}{3\sqrt{5}}& \frac{2}{3}\end{array}\right] \end{gathered}$$

Hence the diagonal matrix is $D=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 9\end{array}\right]$.